Density of hair in left and right armpits and Paired T-Test

New Member

Consider the situation of the density of hairs under the arm pit. A human has two arm pits. My hypothesis is that there is no difference in the density of hairs in the left and right armpits. I recruit 70 subjects and measure the density of hairs. I perform a two way paired T-test to prove the null. I specifically use a paired T-test because the samples come from the left and right air pits of the same subject and I assume that density of hairs in the air pits should be correlated because the hair grows under the same genetics and physiological conditions. The data are found to be highly correlated (r=0.97). Am I incorrect to use a paired T-test ?
Thanks for you help.
Peter

TS Contributor

Consider the situation of the density of hairs under the arm pit. A human has two arm pits. My hypothesis is that there is no difference in the density of hairs in the left and right armpits. I recruit 70 subjects and measure the density of hairs. I perform a two way paired T-test to prove the null. I specifically use a paired T-test because the samples come from the left and right air pits of the same subject and I assume that density of hairs in the air pits should be correlated because the hair grows under the same genetics and physiological conditions. The data are found to be highly correlated (r=0.97). Am I incorrect to use a paired T-test ?
Thanks for you help.
Peter

Given the bold, a t-test is an inappropriate method and will not give you the ability to prove or support the null, irrespective of the test yielding a nonsignificant result. Failure to reject the null does not prove or support the null. You may want to look into an equivalence study.

Otherwise, your general thought process for selecting the paired t-test is correct because the observations are not likely to be independent of one another as you noted.

Super Moderator

I think your choice of a paired t-test makes sense, but do keep in mind here that a non-significant result will not "prove" the null to be true. In general, we cannot use statistical analyses to "prove" things; "proof" implies absolute certainty, and statistical analyses always come with some uncertainty attached. More specifically, a non-significant result simply indicates that you don't have evidence to reject the null; it doesn't mean the null is necessarily true (given that there are always other plausible explanations for a non-significant result, such as imperfect power).

You might want to consider an analytic method that is specifically tailored to cases where the intent is to provide evidence for a null hypothesis. E.g., Bayesian estimation with a ROPE, Bayes factors (exclude the self-plug), or equivalence testing. None of these methods will allow you to prove the null with certainty, but all of them will allow for a slightly more useful conclusion than "I couldn't find evidence to reject the null".

TS Contributor

I think your choice of a paired t-test makes sense, but do keep in mind here that a non-significant result will not "prove" the null to be true. In general, we cannot use statistical analyses to "prove" things; "proof" implies absolute certainty, and statistical analyses always come with some uncertainty attached. More specifically, a non-significant result simply indicates that you don't have evidence to reject the null; it doesn't mean the null is necessarily true (given that there are always other plausible explanations for a non-significant result, such as imperfect power).

You might want to consider an analytic method that is specifically tailored to cases where the intent is to provide evidence for a null hypothesis. E.g., Bayesian estimation with a ROPE, Bayes factors (exclude the self-plug), or equivalence testing. None of these methods will allow you to prove the null with certainty, but all of them will allow for a slightly more useful conclusion than "I couldn't find evidence to reject the null".